Solve 2^frac{2-4b{2}+2b}*b^2-2^2/8b-2a

$$2 ^ { \frac { 2 - 4 b } { 2 } + 2 b } \cdot \frac { b ^ { 2 } - 2 ^ { 2 } } { 8 b - 2 a }$$

Evaluate

$$\frac{b^{2}-4}{4b-a}$$

Short Solution Steps

Divide each term of 2-4b by 2 to get 1-2b.

$$2^{1-2b+2b}\times \frac{b^{2}-2^{2}}{8b-2a}$$

Combine -2b and 2b to get 0.

$$2^{1}\times \frac{b^{2}-2^{2}}{8b-2a}$$

Calculate 2 to the power of 1 and get 2.

$$2\times \frac{b^{2}-2^{2}}{8b-2a}$$

Calculate 2 to the power of 2 and get 4.

$$2\times \frac{b^{2}-4}{8b-2a}$$

Express 2\times \frac{b^{2}-4}{8b-2a} as a single fraction.

$$\frac{2\left(b^{2}-4\right)}{8b-2a}$$

Factor the expressions that are not already factored.

$$\frac{2\left(b-2\right)\left(b+2\right)}{2\left(-a+4b\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{\left(b-2\right)\left(b+2\right)}{-a+4b}$$

Expand the expression.

$$\frac{b^{2}-4}{-a+4b}$$

Factor

$$\frac{\left(b-2\right)\left(b+2\right)}{4b-a}$$